seismic loads lateral load flow frames and shear walls
DESCRIPTION
SEISMIC LOADS LATERAL LOAD FLOW FRAMES and SHEAR WALLS. SEISMIC LOAD. Determine Spectral Response Parameters at design location. At 37.80 N , -122.37 W : Ss = 1.50 S 1 = 0.60. Determine Site Coefficients. Site Class : D Ss > 1.25 Fa = 1.0 S 1 > 0.5 Fv = 1.5. - PowerPoint PPT PresentationTRANSCRIPT
SEISMIC LOADS
LATERAL LOAD FLOW
FRAMES and SHEAR WALLS
SEISMIC LOAD
At 37.80 N , -122.37 W :
Ss = 1.50
S1 = 0.60
Determine Spectral Response Parameters at design location
Site Class : D
Ss > 1.25
Fa = 1.0
S1 > 0.5
Fv = 1.5
Determine Site Coefficients
Determine Design Spectral Acceleration Parameters
SMS = (1.0)(1.5) = 1.5
SDS = (2/3)(1.5) = 1.0
Cs = SDS /(R/I)
=1.0/(R/I)
Class II : I = 1.0
Ordinary Moment Resisting Frame :
R = 3.5
V = 1.0/3.5 W
0.3 W
Seismic Load is generated by the inertia of the mass of the structure : VBASE
Redistributed (based on relative height and weight) to each level as a ‘Point Load’ at the center of mass of the structure or element in question : FX
VBASE Wx hx
(w h)
( VBASE )
(Cs)(W)VBASE =
Fx =
Total Seismic Loading :
VBASE = 0.3 W
W = Wroof + Wsecond
Wroof
Wsecond flr
W = Wroof + Wsecond flr
VBASE
Redistribute Total Seismic Load to each level based on relative height and weight
Fx =
Froof
Fsecond flr
VBASE (wx)(hx)
(w h)
Fx =VBASE (wx)(hx)
(w h)
In order to solve the equivalent lateral force distribution equation, we suggest you break it up into a spreadsheet layout
Floor w h (w)(h) (w)(h)/(w)(h) Vbase Fx
Roof 166.67k 30ft 5000k-ft 0.625 110k68.75k
2nd 200k 15ft 3000k-ft 0.375 110k 41.25k
(366.67k) (8000k-ft) (110k)
Vbase = 0.3W = 0.3(166.67k+200k) = 0.3(366.67k) = 110k
Load Flow to Lateral Resisting System :
Distribution based on Relative Rigidity
Assume Relative Rigidity : Single Bay MF :Rel Rigidity = 1
2 - Bay MF :Rel Rigidity = 2
3 - Bay MF :Rel Rigidity = 3
Distribution based on Relative Rigidity :
R = 1+1+1+1 = 4
Px = ( Rx / R ) (Ptotal)
PMF1 = 1/4 Ptotal
Lateral Load Flow
diaphragm > collectors/drags > frames
STRUCTURAL DIAPHRAGM
A structural diaphragm is a horizontal structural system used to transfer lateral loads to shear walls or frames primarily through in-plane shear stress
Basically, combined with vertical shear walls or frames IT ACTS LIKE A LARGE I-BEAM
STRUCTURAL DIAPHRAGM
Flexible or Semi-flexible Type:
PlywoodMetal Decking
STRUCTURAL DIAPHRAGM
Rigid Diaphragm Type:
Reinforced Concrete SlabConcrete-filled Metal Deck composite SlabBraced/horizontal truss
STRUCTURAL DIAPHRAGM
Rigid Diaphragm:
Almost no deflectionCan transmit loads through torsion
Flexible Diaphragm:
Deflects horizontallyCannot transmit loads through torsion
COLLECTORS and DRAGS
COLLECTORS and DRAG STRUTS
A beam element or line of reinforcement that carries or “collects” loads from a diaphragm and carries them axially to shear walls or frames.
A drag strut or collector behaves like a column.
Lateral Load Flow
diaphragm > collectors/drags > frames
DIAPHRAGM
COLLECTOR
COLLECTOR
FRAME
FRAME
Lateral Load Flow
diaphragm > collectors/drags > frames
DIAPHRAGM
COLLECTOR
COLLECTOR
FRAME
FRAME
LATERAL
LOAD
Lateral Load Flow
diaphragm > collectors/drags > frames
DIAPHRAGM
COLLECTOR
COLLECTOR
FRAME
FRAME
LATERAL
LOAD
DIAPHRAGM
COLLECTOR
COLLECTOR
FRAME
FRAME
LATERAL
LOAD
FRAMECOLLECTOR
COLLECTOR
LATERAL FORCE RESISTING SYSTEMS:
MOMENT Resisting frames
Diagonally BRACED frames
SHEAR walls
INSTABILITY OF THE FRAME
Pinned connectionscannot resist rotation.
This is not a structurebut rather a mechanism.
STABILIZE THE FRAME
FIX ONE OR MORE OF THE BASES
FIX ONE OR MORE OF THE CORNERS
STABILIZE THE FRAME
STABILIZE THE FRAME
ADD A DIAGONAL BRACE
RELATIVE STIFFNESS OF FRAMES AND WALLS
LOW DEFLECTION
HIGH STIFFNESS
ATTRACTS MORE LOAD
HIGH DEFLECTION
LOW STIFFNESS
ATTRACTS LESS LOAD
BRACED FRAMES
BRACED FRAMES
SHEAR WALLS
SHEAR WALLS
SHEAR WALLS
SHEAR WALLS
SHEAR WALLS
MOMENT FRAMES
MOMENT FRAMES
MOMENT FRAMES
INDETERMINATE STRUCTURES
SOLVE BY “PORTAL FRAME METHOD”
MOMENT FRAMES
SOLVE BY “PORTAL FRAME METHOD”
PINNED BASE =4 UNKNOWNS, 3 EQUATIONS, STATICALLY INDETERMINATE TO FIRST DEGREE
MOMENT FRAMES
SOLVE BY “PORTAL FRAME METHOD”
FIXED BASE =6 UNKNOWNS, 3 AVAILABLE EQUATIONS OF EQUILIBRIUMSTATICALLY INDETERMINATE TO THE 3RD DEGREE